Using properties of determinants, prove that:

$\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$

If the determinant $\left| {\begin{array}{*{20}{c}}{a\, + \,p}&{1\, + \,x}&{u\, + \,f}\\ {b\, + \,q}&{m\, + \,y}&{v\, + \,g}\\{c\, + \,r}&{n\, + \,z}&{w\, + \,h}\end{array}} \right|$ splits into exactly $K$ determinants of order $3$, each element of which contains only one term, then the value of $K$, is

Let $P=\left[\begin{array}{ccc}1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1\end{array}\right]$ and $I$ be the identity matrix of order $3$ . If $\left.Q=q_{i j}\right]$ is a matrix such that $P^{50}-Q=I$, then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

If ${D_r} = \left| {\begin{array}{*{20}{c}}{{2^{r – 1}}}&{{{2.3}^{r – 1}}}&{{{4.5}^{r – 1}}}\\x&y&z\\{{2^n} – 1}&{{3^n} – 1}&{{5^n} – 1}\end{array}} \right|$, then the value of $\sum\limits_{r = 1}^n {{D_r} = } $

$\left| {\begin{array}{*{20}{c}}{1 + {{\sin }^2}\theta }&{{{\sin }^2}\theta }&{{{\sin }^2}\theta }\\{{{\cos }^2}\theta }&{1 + {{\cos }^2}\theta }&{{{\cos }^2}\theta }\\{4\sin 4\theta }&{4\sin 4\theta }&{1 + 4\sin 4\theta }\end{array}} \right| = 0$ then $\sin \,4\theta $ equal to